Differential Equations: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Differential equations

Exponential growth/decay: dy/dt = ky → y = y₀e^kt
Half-life: k = ln 2 / t_½
Newton's law of cooling: T(t) = T_a + (T₀ − T_a)e^−kt
Integrating factor: y′ + p(x)y = q(x): μ = e^{∫p dx}, (μy)′ = μq
Mixing tank: dm/dt = r·c − (r/V)·m

Integrals

Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
Definite integral: ∫_a^b f dx = F(b) − F(a)
Common integrals: ∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves: ∫_a^b (f − g) dx

Method

Rewrite the integrand as a sum of simple terms where possible.
Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for e^x, 1/x, sin and cos.
Use substitution when you see a function and its derivative together.
For a definite integral, evaluate F(b) − F(a).

Derivatives

Power rule: (xⁿ)′ = n·xⁿ⁻¹
Product rule: (uv)′ = u′v + uv′
Quotient rule: (u/v)′ = (u′v − uv′)/v²
Chain rule: f(g(x))′ = f′(g(x))·g′(x)
Common derivatives: (eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x, (cos x)′ = −sin x

Method

Identify the structure: sum, product, quotient or composition.
Apply the power rule to each power of x.
Use the product, quotient or chain rule where needed.
Simplify, and evaluate at the point if asked.

Numerical methods

Bisection: c = (a + b)/2, width after n steps (b − a)/2ⁿ
Newton–Raphson: x_n+1 = xₙ − f(xₙ)/f′(xₙ)
Linear interpolation: y = y₁ + (y₂ − y₁)(x − x₁)/(x₂ − x₁)
Trapezoid rule: T = (h/2)[f₀ + 2f₁ + … + 2f_n−1 + fₙ], h = (b − a)/n
Simpson’s rule (n even): S = (h/3)[f₀ + 4f₁ + 2f₂ + … + 4f_n−1 + fₙ]
Trapezoid error bound: |E| ≤ (b − a)h²M/12, M = max|f″|
Euler: y_n+1 = yₙ + h·f(tₙ, yₙ)
Heun: y_n+1 = yₙ + (h/2)(k₁ + k₂), k₂ = f(tₙ + h, yₙ + hk₁)
RK4: y_n+1 = yₙ + (h/6)(k₁ + 2k₂ + 2k₃ + k₄)

Method

For a root: check the sign change for bisection, or compute f and f′ for Newton, and keep six or more digits in a table of iterates.
For a table value: interpolate linearly between the two neighbouring points.
For an integral: find h = (b − a)/n, list f at every node, apply the weights (1, 2, …, 2, 1 for trapezoid; 1, 4, 2, …, 4, 1 for Simpson with even n) and multiply by h/2 or h/3.
For an ODE: step from t₀ with yₙ₊₁ = yₙ + h·f(tₙ, yₙ) (Euler) or the averaged slope (Heun), one row per step, until you reach the requested t.

2Practice problems

1.Differential equationsMedium

A can of soda at 3 °C is taken out of the fridge into a room at 23 °C. After 10 min its temperature is 11 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). How long after the start, t, does it take to reach 8 °C?

T₀ = 3 °CT_a = 23 °CT(10) = 11 °CT = 8 °C

t = min

2.Differential equationsHard

A tank holds 400 L of pure water. Brine with 0.1 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 95 min?

V = 400 Lm₀ = 0 kgc = 0.1 kg/Lr = 2 L/min

m = kg

3.IntegralsMedium

Evaluate I = ∫₀² 2x·(x² + 1)¹ dx.

f(x) = 2x·(x² + 1)¹[0, 2]

I =

4.IntegralsHard

The region between the line y = 2x and the parabola y = 0.5x² is rotated about the x-axis. Find the volume V of the solid.

y = 2xy = 0.5x²

V =

5.DerivativesMedium

Let f(x) = x·sin(2x), with x in radians. Find f′(2).

f(x) = x·sin(2x)x = 2 rad

f′(2) =

6.DerivativesHard

Let f(x) = x³ − 12x² + 36x − 3. Find the x-coordinate x of the local minimum.

f(x) = x³ − 12x² + 36x − 3

x =

7.Numerical methodsMedium

Approximate ∫₀² 1/(1 + x³) dx with the trapezoid rule using n = 4 strips. Give T.

f(x) = 1/(1 + x³)[a, b] = [0, 2]n = 4

T =

8.Numerical methodsHard

Take one step of the classical Runge–Kutta method (RK4) with h = 0.1 for y′ = 1 + sin(ty), y(0) = 1. Find y₁ ≈ y(0.1).

y′ = 1 + sin(ty)y(0) = 1h = 0.1

y₁ =

3Solutions

1.Differential equations

Answer: t = 5.632 min

First-order differential equations: dy/dt = k·y has the solution y = y₀·e^kt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e^−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e^{∫p dx}: (μy)′ = μq, so μy = ∫μq dx + C.

Cooling constant

k = ln(-20/-12)/10 = 0.05108 min⁻¹

Time to reach the target

t = ln(-20/(−15))/0.05108 = 5.632 min

Answer

t = 5.632 min

2.Differential equations

Answer: m = 15.12 kg

Set up the equation

dm/dt = 2·0.1 − (2/400)·m = 0.2 − 0.005·m

Solve

The integrating factor is e^rt/V; the solution approaches the equilibrium c·V.

m(t) = 40 + (−40)·e^−t/200

Substitute t

m(95) = 40 + (−40)·0.6219 = 15.12 kg

Answer

m = 15.12 kg

3.Integrals

Answer: I = 12

Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.

Substitution

u = x² + 1, du = 2x dx

I = ∫₁⁵ u¹ du

Integrate and evaluate

I = [u²/2]₁⁵ = (25 − 1)/2 = 12

Answer

I = 12

4.Integrals

Answer: V = 107.2

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Intersection points

2x = 0.5x² → x = 0, x = 4

Washer method

V = π∫₀⁴ ((2x)² − (0.5x²)²) dx

Integrate

V = π(4·4³/3 − 0.25·4⁵/5) = π·34.13 = 107.2

Answer

V = 107.2

5.Derivatives

Answer: f′(2) = −3.371

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Product and chain rule

f′(x) = sin(2x) + 2x·cos(2x)

Substitute

f′(2) = −0.7568 + 4·(−0.6536) = −3.371

Answer

f′(2) = −3.371

6.Derivatives

Answer: x = 6

Derivative

f′(x) = 3x² − 24x + 36 = 3·(x² − 8x + 12)

Solve f′(x) = 0

x = 2 or x = 6

Which one is which

f″(x) = 6x − 24: negative at a maximum, positive at a minimum.

f″(2) = −12, f″(6) = 12

x = 6

Answer

x = 6

7.Numerical methods

Answer: T = 1.087

Numerical integration: split [a, b] into n strips of width h = (b − a)/n with nodes x₀ … xₙ. Trapezoid rule: T = (h/2)[f₀ + 2f₁ + 2f₂ + … + 2f_n−1 + fₙ]. Simpson's rule (n even): S = (h/3)[f₀ + 4f₁ + 2f₂ + 4f₃ + … + 4f_n−1 + fₙ]. The trapezoid error shrinks like h² (bound (b − a)h²·M/12 with M = max|f″|), Simpson's like h⁴, so Simpson is usually far more accurate for the same nodes.

Nodes and values

h = 0.5

Σ wᵢfᵢ = 1 + 2·0.888889 + 2·0.5 + 2·0.228571 + 0.111111 = 4.346032

Trapezoid rule

T = (0.5/2)·4.346032 = 1.087

Answer

T = 1.087

This integral has no elementary antiderivative, which is exactly when a numerical rule is needed.

8.Numerical methods

Answer: y₁ = 1.105

Numerical solution of y′ = f(t, y): Euler's method steps along the slope at the start of each step: y_n+1 = yₙ + h·f(tₙ, yₙ). Heun's method (improved Euler) averages the slope at the start and at an Euler prediction: k₁ = f(tₙ, yₙ), k₂ = f(tₙ + h, yₙ + h·k₁), y_n+1 = yₙ + (h/2)(k₁ + k₂). The classical Runge–Kutta method (RK4) combines four slopes: y_n+1 = yₙ + (h/6)(k₁ + 2k₂ + 2k₃ + k₄). The global error of Euler is proportional to h, Heun's to h², RK4's to h⁴.

Four slopes

k₁ = f(0, 1) = 1

k₂ = f(0.05, 1.05) = 1.052476

k₃ = f(0.05, 1.052624) = 1.052607

k₄ = f(0.1, 1.105261) = 1.110301

Combine

y₁ = 1 + (0.1/6)(1 + 2·1.052476 + 2·1.052607 + 1.110301) = 1.105

Answer

y₁ = 1.105

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